I am having trouble with the following problem:
Do there exist polynomials $a(x), b(x), c(y), d(y) \in \mathbb{R}[x]$ for which the following holds? $$ \forall x, y \qquad 1+xy+x^2y^2 = a(x)d(y) - b(x)c(y) $$
I do not see how to approach this problem -- should I start making guesses for $a,b,c,d$?
Suppose that $\, f(x,y) := 1 + xy + x^2y^2 .\,$ The function given by $\, g(x,y) := a(x)d(y)-b(x)c(y) \,$ has rank at most $\,2,\,$ by definition, which implies that the determinant of the $\,3\times 3\,$ matrix $\, \{g(x_i,y_j)\}_{i,j=1}^3 \,$ for any $\, x_i, y_j \,$ is $\,0.\,$ This is a direct computation.
Let us call any function of two variables of the form $\, u(x)v(y) \,$ a separable or rank $1$ function. Any linear combination $\, g(x,y) := \sum_{n=1}^r a_n\,u_n(x)v_n(y) \,$ of separable functions for any $\, u_n, v_n \,$ has rank at most $\,r.\,$ This is because the $\, n\times n\,$ matrix $\, \{g(x_i,y_j)\}_{i,j=1}^n \,$ for any $\, x_i, y_j \,$ has rank at most $\,r.\,$
In the case of $\,f(x,y),\,$ the determinant of the matrix $\, \{f(i,j)\}_{i,j=1}^3 \,$ is $\,4.\,$ Therefore there do not exist functions $\, a,b,c,d \,$ such that $\, \forall x, y \qquad f(x,y) = g(x,y). \,$ Note that $\, f(x,y) \,$ is the sum of $3$ separable functions and has rank exactly $3.$